Question

At the foot of the mountain, the angle of elevation of its summit is $45^{o},$ after ascending $1000m$ towards the mountain up a slope of $30^{o}$ inclination, the angle of elevation is found to be $60^{o}$ . The height (in $km$ ) of the mountain is (take $\sqrt{3}=1.7$ )

Answer 1

Let $P$ be the summit of the mountain and $Q$ be the foot.
Let $A$ be the first position and $B$ the second position of observation.
$BN$ and $BM$ are $\bot$ s from $B$ to $PQ$ and $AQ$ respectively.
Then, $AB=1000m=1km$ ,
$\angle MAB=30^{o},$
$\angle MAP=45^{o},$
$\angle NBP=60^{o}$
Now, $\angle BAP=\angle MAP-\angle MAB=45^{o}-30^{o}=15^{o}$
$\angle APB=\angle APN-\angle BPN=45^{o}-30^{o}=15^{o}$
$\therefore \Delta ABP$ is isosceles
$\therefore AB=BP$
But, $AB=1$ kilometre
$\therefore BP=1$ kilometre.
Now, $PQ=PN+NQ=PN+BM$
$=BPsin 60^{o}+ABsin ⁡ 30^{o}$
$=1\cdot \frac{\sqrt{3}}{2}+1\cdot \frac{1}{2}=\frac{\sqrt{3} + 1}{2}=1.35k\text{m}$